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iosarraysswiftfor-loopswift3
Joe
Joe

Reputation: 1968

How do I write a for-loop in Swift 3 for an array that I modify during the for loop?

So, I have a for-loop that looks similar to this:

for var i = 0; i < results.count ; i += 1 {
   if (results[i] < 5) {
      results.removeAtIndex(i)
      i -= 1
   }
}

This used to work. But when I changed it to the preferred Swift 3.0 syntax:

for var i in 0..<results.count {
   if (results[i] < 5) {
      results.removeAtIndex(i)
      i -= 1
   }
}

I get an array IOOBE exception because it doesn't re-check the count and continues on until the original results.count.

How do I fix this? It works now, but I don't want to get into trouble in the future.

Upvotes: 55

Views: 137686

Answers (4)

Brandon Shega
Brandon Shega

Reputation: 769

Could you use a filter instead?

let numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
let greaterThan4 = numbers.filter{$0 >= 5}
print(greaterThan4)

Upvotes: 19

Unheilig
Unheilig

Reputation: 16292

While the solution making use of filter is a fine solution and it's more Swift-ly, there is another way, if making use of for-in is, nonetheless, still desired:

func removeBelow(value: Int) {
    var results = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

    for i in (0 ..< results.count).reversed() {
        if (results[i] < value) {
            results.remove(at: i)
        }
    }

    print(results)
}

removeBelow(value: 5)

Result:

[5, 6, 7, 8, 9, 10]

The problem with removeAtIndex within the loop is that it will not cause the array to re-index itself in-place and thus causing an array out of bounds exception due to count not being updated.

By traversing backwards, the out of bounds exception can thus be avoided.

Upvotes: 86

Miguel Herrero
Miguel Herrero

Reputation: 39

If your loop goes forward...

for var i in (0..<results.count) where results.indices.contains(i) { 

//if the index doesn't exist, the loop will be stopped.

if (results[i] < 5) {
        results.removeAtIndex(i) 
    }

}

Upvotes: 1

Luka Jacobowitz
Luka Jacobowitz

Reputation: 23512

If you want to continue using a for-loop, you can enumerate over both index and element using enumerate:

for (index, element) in results.enumerate() {
   if (element < 5) {
     results.removeAtIndex(index)
   }
}

Although depending on what you're doing in your loop, the filter method might be a better idea.

Upvotes: 15

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